Aptitude - Boats and Streams - Discussion

There is a formula.

B/S = (T1 + T2) / (T1 ~ T2).
Here,

* B is the speed of boat in still water.
* S is the speed of stream/current/river.
* T1 is the time taken in upstream/ downstream.
* T2 is the time taken in downstream/ upstream.

Given:

Twice as long to row ' against' the stream (is. Upstream) as to row same distance in 'favor' of the stream(ie. downstream).

Solution:

Upstream : downstream
Time: 2 : 1 ( T1 = 2 *T2 )
So, T1 = 2 and T2 = 1.

Apply in the formula,
B/S = (2+1)/(2-1) = 3/1.
B:S = 3:1.

Let time taken to travel against the stream is 2,

Time taken to travel in favour of stream is 1,

As we know D=S*T,

As Distance is same (constant) so we can neglect it,

Speed downstream (Sd) = D/t =>1/2-----take it as a,

Speed upstream (Su) = D/t =>1-----take it as b,

We need to find ratio of the speed of the boat (in still water) to the stream is, i.e.

1/2(a+b):1/2(a-b).

==>1/2(1+1/2):1/2(1-1/2).

==>3:1.

If the math is delivered in question as MCQ then it can be solved in a very easy and quick process:.

Let downstream speed = 80.

So, upstream = 40.

So, Boat speed, 60 + current speed 20 = 80.

Boat speed, 60 - current speed 20 = 40,

So ratio = 60 : 20 = 3 : 1, it's informal but by letting this type of numbers we can solve MCQ within a short period.

Let the speed of man in still water = M.

Let the speed of stream = S.

Let speed of man in direction of stream = X.

Therefore,

M + S = X (Man in direction of stream).

M - S = X/2 (Man takes twice as much time it means his speed is 1/2*X).

-------------.

2M = 1.5*X.
=> M = 0.75*X.
=> S = 0.25*X.
=> M/S = 0.75/0.25 = 3/1.

Let's call Speed of Boat as B and Speed of Stream as S.
Distance = Time x Velocity.
Let's call Time with T.

Distance with Favour is (B + S) x T.
Distance against the stream is (B - S) x T.


Since the distance is twice if the stream is in favour
So, (B + S) x T = (B - S) x T x 2,
B + S = 2B - 2S,
B = 3S.
So, the ratio is 3 : 1.

Let d= distance and x = time.
Speed in still water = 1/2(speed in down stream + speed in up stream).
= 1/2 (d/x + d/2x) => eqn 1.
Rate of stream = 1/2( speed in down stream - speed in up stream).
= 1/2(d/x - d/2x) => eqn 2;
We need eqn 1 : eqn 2.

On simplification, we will get 3:1.

Let speed of boat = b.
Speed of current = c.
Upstream spped = b - c.
Downstream speed= b + c.

Since distance is constant, and time taken in upward is twice the time taken in downstream.
So, 2 * ((d) / (b+c)) = (d) / (b - c).

After solving it.
2b - 2c = b + c.
b/c = 3/1.

Dear @Srikanth.

In favour of stream means 'downstream' ==> (sd) = 1/1---taking it as a,

Against the stream means 'upstream' ==> (Su) = 1/2---taking it as b, now speed of stream will be positive value ((Sd)-(Su))/2.

Am I correct here reply.

Suppose speed upstream is x kmph.

Speed of current y kmph.

Time suppose is = t.

So (x+y)*t = (x-y)*2t because twice time taken from here we get.

y = x/3.

Now we need.

(x+y)/2:(x-y)/2.

Put value of why we get 2:1.

I think its right answer.

Let speed of boat in downstream be 'x' kmph then speed of boat in upstream is 'x/2' kmph because time taken is twice hence.

A be the speed of boat and b be the speed of stream.
A+B = B.
A-B = X/2.
Hence A=3/4 and B=1/4.

Ratio is 3:1.